Is it possible to have a nonsplit short exact sequences of vector bundles (on some smooth variety) $0\rightarrow V_1 \rightarrow V_2 \rightarrow V_3 \rightarrow 0$. Such that $V_2\cong V_1\oplus V_3$ as vector bundles?

1$\begingroup$ What do you mean by 'as objects'? $\endgroup$– JefNov 17 '20 at 18:49

2$\begingroup$ As vector bundles. Not as extensions (which obviously is impossible according to the assumptions). $\endgroup$– user127776Nov 17 '20 at 18:50

1$\begingroup$ See also mathoverflow.net/questions/163041/… . $\endgroup$– LSpiceNov 17 '20 at 19:24

4$\begingroup$ You can take a nonsplit short exact sequence on $\mathbb{P}^1$ of the form $$ 0 \to \mathcal{O} \to \mathcal{O} \oplus \mathcal{O}(n) \to \mathcal{O}(n) \to 0.$$ It exists as soon as $n \geq 2$, as one can check by computing the Ext$^1$ group. $\endgroup$– Francesco PolizziNov 17 '20 at 19:26

5$\begingroup$ @FrancescoPolizzi : but for example, there is a nonsplit sequence $0 \to \mathcal O \to \mathcal O(1)^{\oplus 2} \to \mathcal O(2) \to 0$ corresponding to a nonzero element in $Ext^1(\mathcal O(2), \mathcal O)$. So given a nonzero class in $Ext^1(\mathcal O, \mathcal O(n))$ do you know that the middle term will be given by $\mathcal O \oplus \mathcal O(n)$ ? $\endgroup$– Nicolas HemelsoetNov 17 '20 at 20:06
$\newcommand{\cO}{\mathcal{O}}$Consider exact sequence of trivial vector bundles $$0\to\cO\xrightarrow{\left(\begin{matrix}x \\ y\end{matrix}\right)}\cO\oplus\cO\xrightarrow{\left(\begin{matrix}y & x\end{matrix}\right)}\cO\to 0$$ on $X=\mathbb{A}^2_{x,y}\setminus\{0\}$. One checks easily that it is exact on stalks (but the same sequence on $\mathbb{A}^2$ is not exact at $(x,y)=(0,0)$). It is the pullback of $0\to\cO(1)\to \cO\oplus\cO\to \cO(1)\to 0$ from $\mathbb{P}^1$. A splitting of this sequence would be a pair of functions $f_1,f_2\in H^0(\mathbb{A}^2\setminus\{0\},\cO)=k[x,y]$ such that $yf_1xf_2=1$ but there is no such pair.
On the positive side, any such sequence has to be split if $X$ is proper. In this case the spaces $Hom(V_3, V_1)$ and $Hom(V_3, V_3)$ are finitedimensional over $k$. Applying $Hom(V_3,)$ to this exact sequence we get a left exact sequence of finitedimensional vector spaces $$0\to Hom(V_3, V_1)\to Hom(V_3,V_2)\to Hom(V_3,V_3)$$ The dimension of the vector space in the middle $Hom(V_3,V_2)\simeq Hom(V_3, V_1\oplus V_3)$ is equal to the sum of the dimensions of first and third terms. Therefore, the sequence has to be exact on the right and, in particular, the identity $Id_{V_3}\in Hom(V_3, V_3)$ lifts to a morphism $Hom(V_3, V_2)$ that gives a splitting.

$\begingroup$ Yes my $X$ is proper. Your argument is supersmart. So there are no examples of such thing on projective line. This refutes one of the comments above. Thanks. $\endgroup$ Nov 17 '20 at 20:11

1$\begingroup$ @user127776 : I think the explanation is given by my comment below $\endgroup$ Nov 17 '20 at 20:11

$\begingroup$ Does this really refute my example? My middle term is not balanced, how do you reduce yourself to the case of the sum of two trivial bundles? $\endgroup$ Nov 17 '20 at 20:37

1$\begingroup$ The explanation given above for proper variety is general. It is not specialized to the specific short exact sequence of trivial bundles given at the start. $\endgroup$ Nov 17 '20 at 21:00

$\begingroup$ Ok, I see. I was confused by "any such sequence" at the beginning of the second part of the answer. $\endgroup$ Nov 17 '20 at 21:50